The first index need not be $1\;$ and may be an arbitrary integer (see an example where the difference in starting indices was concealed leading to a curious observation). When there is no possibility of confusion the first index is omitted all together:

This sequence may or may not have a limit. If the limit $\displaystyle\lim_{i\rightarrow\infty}s_{i}\;$ exists then the series is said to converge, or to be convergent. In this case the expression $\displaystyle\sum_{i=1}^{\infty} a_{i}\;$ is assigned a value, the limit of the sequence $s_{i}.\;$ The series that do not converge are said to diverge, or to be divergent. Several examples of convergent and divergent series are available elsewhere.

Here we focus on convergence properties of the harmonic series, a series of the reciprocals of positive integers $\displaystyle a_{n} = \frac{1}{n}.$

The partial sums of the harmonic grow without bound which, in particular, means that the harmonic series is divergent. There are many proofs of that result. Here is one that is credited to the famous philosopher of the Middle Ages, Nicolas Oresme. It is clear that a series with constant terms, however small, is divergent. Because the partial sums do grow without bound. For the harmonic series we can write

where the first sign of equality "=" tells us that series in fact remained the same although its terms have been grouped in a peculiar way. The last equality sign "=" only meant to indicate that we get the same series but with the terms in a simpler form.

Thus the sequence of partial sums of the harmonic series exceeds term-by-term the sequence of partial sums of a series that diverges to infinity. So, the same can be said of the harmonic series as well.

A recent proof due to Leonard Gillman starts with a contrary assumption that the series \sum 1/n converges to a finite number S:

with the conclusion that $S\gt S,\;$ which is absurd. (Additional details can be found on a separate page.)

Euler has shown that the partial sums of the harmonic series change slowly:

$s_{n} \approx \ln n + \gamma,$

where $\ln n\;$ is a natural logarithm of $n\;$ and $\gamma\;$ is a constant that now bears Euler's name:

$\gamma\approx 0.57721566490153286061\ldots$

(It is not known yet whether Euler's constant is transcendental or, for that matter, if it's irrational. Sometimes it is also referred to as the Euler-Mascheroni constant.)

The series of the reciprocals of all the natural numbers - the harmonic series - diverges to infinity. There are many ways to thin the series as to leave a convergent part. For example, if we leave only the reciprocals of the squares, $\displaystyle\sum_{n\ge 1}\frac{1}{n^{2}},\;$ the series will converge. (This is because the reciprocal of a square, say, $\displaystyle\frac{1}{k^{2}},\;$ is bounded from above by a term $\displaystyle\frac{1}{k(k - 1)}\;$ of the convergent telescoping series.) On the other hand, if we remove the reciprocals of all the composite numbers leaving only those of the primes, the remaining series will still be divergent.

In a 1914 paper, A. J. Kempner proved a surprising result that a removal of the reciprocals of all the integers whose decimal representation contains a specified digit (say 3) leaves a convergent series. The result was strengthened shortly thereafter by F. Irwin in the following form:

Theorem (Irwin, 1916)

If we strike out from the harmonic series those terms whose denominators contain the digit $9\;$ at least $a\;$ times, and, at the same time, the digit $8\;$ at least $b\;$ times, the digit $7\;$ at least $c\;$ times, and so on, to the digit $0\;$ at least $j\;$ times $(a, b, c, \ldots, j\;$ being any given integers), the series so obtained will converge.

(It should be noted that an increase of any of the $10\;$ numbers $a, b, c, \ldots, j\;$ shrinks the set of the removed terms.)

Some eighty years later, H. Behforooz proved a result concerning the density of the terms in the harmonic series that form a convergent series on their own.

Theorem (Behforooz, 1995)

Suppose that $C\;$ is a subset of positive integers and $\displaystyle\sum_{n\in C}\frac{l}{n}\;$ is convergent. For any positive integer $k,\;$ let $N_{k}\;$ be the number of elements in $C\;$ that are $\le 10^{k},\;$ and $M_{k} = 10^{k} - N_{k}.\;$ Then $\displaystyle lim_{k\rightarrow\infty}\frac{N_{k}}{M_{k}} = 0.$

Let's prove Kempner's original theorem.

Theorem (Kempner, 1914)

If the denominators do not include all natural numbers $1, 2, 3, \ldots\;$ but only those numbers which do not contain any figure $9,\;$ the series converges. The method of proof holds unchanged if, instead of $\,9,\;$ any other figure $1, 2, \ldots, 8\;$ is excluded, but not for the figure $0.$

(The proof can be found in more accessible publication, see, e.g., Honsberger, Boas, or Hardy and Wright. Removing digit $0\;$ requires only a slight modification.)

Proof

We split the series of the (remaining) reciprocals into groups with indices k satisfying

$10^{n} \le k \lt 10^{n + 1}, n = 0, 1, 2, \ldots$

The sum of the terms in each group is denoted $a_{n}\;$ so that the series is replaced with

$T = a_{0} + a_{1} + \ldots + a_{n} + \ldots$

We need to estimate the sums $a_{n}.\;$ As we saw in another discussion, there are at most $9^{n+1}\;$ integers between $10^{n}\;$ and $10^{n + 1}\;$ that miss a particular digit. (We consider $n + 1\;$ letter strings from an alphabet of $9\;$ letters. Not all such strings correspond to valid decimal representations of integers.)

On the other hand, the terms in the $\displaystyle\sum_{n\ge 1}a_{n}\;$ are bounded from above by $10^{-n}.\;$ It follows that

The sum $T\;$ is bounded from above implying the convergence of the series.

The assertion of the theorem may be surprising at first sight as the relevant assertion that "all integers have digit $3\;$ in their decimal representation". On second thought, as R. Honsberger has observed, "most" natural numbers contain millions of digits. For the large numbers it will be more surprising not to contain digit 3.